Southern  Branch 
of  the 

University  of  California 

Los  Angeles 


This  book  is  DUE  on  the  last 


Form  L-9-157tt-8,'26 


KEY 


NATURAL  PHILOSOPHY. 


BY 

ISAAC  SHARPLESS,  Sc.D., 

PRESIDENT   OP    HAVEKFORD   COLLEGE, 

AND 

GEO.  MORRIS   PHILIPS,  PH.D., 

PRINCIPAL   OF   STATE    NORMAL    SCHOOL,    WEST   CHESTER,    PA. 


GTATEHOBMALSCHOOL 

1.06  AWOttfES.  -•-,.», 


PHILADELPHIA: 

J.   B.   LIPPINCOTT    COMPANY. 

1893. 


Copyright,  1884,  by  J.  B.  LIPPINCOTT  &  Co 


Copyright,  1893,  by  J.  B.  LIPPINCOTT  COMPANY. 


749  1 


. 

.c  2  3 

5  $3 


PREFACE. 


THIS  Key  is  written  for  the  convenience  of  teachers, — 
to  save  them  the  time  necessary  to  work  out  all  the 
Exercises  of  the  Philosophy.  It  will  assist  them  also 
in  determining  which  of  these  Exercises  to  omit.  While 
all  of  them  will  be  useful  for  a  class  of  good  or  well- 
developed  students,  it  is  undesirable  to  exact  them  all 
of  the  weaker  ones:  in  general,  the  authors  think  it 
unwise  to  require  any  which  the  average  member  of  the 
class  cannot  work  out  for  himself,  at  least  with  a  few 
hints  from  the  teacher.  Properly  used,  anything  which 
requires  original  thinking  of  the  students  is  the  most 
beneficial  part  of  a  treatise;  but  examples  too  difficult, 
which  have  to  be  fully  explained  by  the  teacher,  have 
no  especial  value. 

We  think  these  considerations  will  be  a  guide  to  the 
teacher  in  the  selection  of  the  exercises  which  he  assigns 
to  his  students. 


KEY  TO  NATURAL  PHILOSOPHY. 


fage  17. 

Questions. — 1.  It  is  doubled.  2.  It  is  quadrupled.  3.  It 
is  reduced  to  one-fourth.  4.  It  is  not  changed. 

Page  19. 

Exercise  1.  Matter  is  never  destroyed.  "Water  changes  to 
invisible  vapor;  gunpowder  to  invisible  gases;  house-gas 
combines  with  oxygen  in  the  air  and  forms  new  substances. 

2.  Blotting-pads  owe  their  utility  to  porosity;   rubber 
bands,  to  elasticity;  watch-springs,  to  elasticity;  pop-guns, 
to  elasticity  of  air;  putty,  to  adhesion ;  hammers,  to  inertia ; 
piano-strings,  to  elasticity  ;  water-filters,  to  porosity. 

3.  Because  the  sugar  goes  between  the  pores  of  the  water. 

4.  Inertia. 

5.  Because  the  little  air  in  it  tends  to  expand,  and  the 
outward  pressure  is  not  resisted.     It  collapses  because  the 
air  leaks  out  through  the  pores. 

6.  The  divisibility  of  matter  into  very  minute  portions. 

7.  By  seeing  which  would  scratch  the  other. 

8.  The  molecules  are  separated.    The  mass  is  unchanged  ; 
the  density  diminished  ;  the  weight  unchanged  ;  the  volume 
increased. 

9.  Water. 

10.  Divide  the  mass  by  the  volume.     Divide  the  mass  by 
the  density. 

11.  Graphite,  or  the  "  lead"  of  lead-pencils. 

12.  Because  the  earth  pulls  all  the  particles  of  each  the 
same,  and  draws  each  particle  just  as  if  it  were  not  con- 
nected with  the  others. 

1*  6 


6  KEY  TO  NATURAL   PHILOSOPHY. 

13.  It  is  three  times  as  far  from  the  centre  of  the  earth, 
and,  as  attraction  varies  inversely  as  the  square  of  the  dis- 
tance, it  will  weigh  one-ninth  of  its  weight  at  the  surface. 

14.  One  thirty-six-hundredth  of  its  weight  at  the  surface, 
or  -jgY^r  —  Tff  °f  a  Pound- 

15.  Since  the  volumes  of  spheres  are  as  the  cubes  of  their 
radii,  the  volume  would   be  one-eighth  the  volume  of  the 
earth  ;  and,  as  they  have  the  same  density,  the  mass  would 
be  one-eighth  also.    Hence  at  the  same  distance  the  attrac- 
tion would  be  one-eighth.     But,  the  body  being  only  half  as 
far  from  the  centre,  the  attraction  would  be  increased  four- 
fold from  the  effect  of  nearness,  if  the  mass  remained  the 
same.     Both  causes  acting,  the  weight  would  be  100  X  i  X 
4  =  50. 

Page  2O. 

Questions. — Train  starting — accelerated.  Train  stop- 
ping— retarded.  Ball  upward — retarded.  Ball  falling — 
accelerated.  Hands  of  a  watch — uniform.  Kiver  and 
winds — varied. 

Page  23. 

Question.  Momenta  equal.     Velocities  as  2  to  1. 

Page  26. 

Exercise  2.  16  X  200  =  3200  grammetres.  3200  -=-  1000 
=  3.2  kilogrammetres  (Ans.). 

Page  28. 

Exercise  L  25  pounds  per  square  inch,  multiplied  by  165, 
the  number  of  square  inches,  gives  4125  pounds'  pressure 
on  the  piston.  This,  multiplied  by  4,  the  number  of  feet 
travelled  by  the  piston  in  each  revolution  of  the  engine, 
and  150,  the  number  of  revolutions  per  minute,  gives 
2,475,000  foot-pounds  per  minute.  2,475,000  -=-  33,000  = 
75  horse-power. 

2.  20  X  10  =  200  foot-pounds  (Ans.}. 


KEY  TO  NATURAL  PHILOSOPHY. 


3.  2  tons  =  4000  pounds.     4000  X  600  =  2,400,000  foot- 
pounds  per  minute.      2,400,000  -=-  33,000  =  72^   horse- 
power (Ans.}. 

4.  72^-  -T-  2  =  36^.  horse-power  required  to  do  the  work 
in  2  minutes. 

72^r  -=-  5  =  14^-  horse-power  to  do  it  in  5  minutes. 

72j8T  -=-  i  =  145^-  horse-power  to  do  it  in  one-half  minute. 

Question  (Art.  71).  As  there  are  980  ergs  in  1  gram,  1000 
grams  in  a  kilogram,  and  100  centimetres  in  a  metre,  1  kilo- 
grammetre  =  980  X  1000  X  100  =  98,000,000  ergs  (Ans.). 

Page  34. 

1.  It  will  slide  when  the  line  downward  from  the  centre 
of  gravity  falls  within  the  base  of  support,  and  roll  when 
it  falls  outside. 

2.  To  bring  the  centre  of  gravity  in  front  of  the  vertical 
line  through  the  feet,  so  that  our  weight  will  assist  the 
muscles  in  rising. 

3.  Because  the  stick  enables  us  more  quickly  to  place  the 
centre  of  gravity  directly  over  the  fence  when  it  gets  a 
little  out. 

4.  When  it  is  at  the  bottom  of  its  swing. 

5.  A  cone  on  its  apex  is  in  unstable  equilibrium  ;  on  its 
base,  in  stable ;  on  its  side,  in  neutral. 

6.  Because  when  he  leans  forward  he  brings  his  centre 
of  gravity  in  front  of  his  base  of  support,  and  he  cannot 
place  any  part  of  himself  backward  to  balance  it.     Hence 
he  cannot  rise. 

7.  In  both  cases  it  should  be  low,  to  avoid  the  danger  of 
overturning. 

8.  Because  in  one  case  the  centre  of  gravity  hangs  on 
the  nail,  in  the  other  it  is  balanced  upon  it.     In  one  case 
the  equilibrium  is  stable,  in  the  other  unstable. 

9.  The  man,  being  only  one-fourth  as  far  from  the  centre 
of  the  earth,  and  inside  of  it,  would  (Art.  41)  weigh  only 
one-fourth  as  much. 


KEY  TO  NATURAL   PHILOSOPHY. 


Pages  35,  36. 

Exercise  1.  s  =  %gt*. 

g  =  32.2,  \  g  =  16.1,  t  =  5,  ?2  =  25. 

Hence  space  =  16.1  X  25  =  402.5  feet  (Ans). 

2.  16.1  X  W=  16.1  X  42^  =  680.225  feet  (Aw). 

3.  Here  ^  =  9.8  metres. 

Hence  s  =  4.9  X  4  =  19.6  metres  (Aw). 

5.  Here  s  =  300  metres. 

Hence  t3  =  300  -=-  4.9  =  61.2245,  and  t  =  1/61.2245 
7.82  seconds  (Aw). 

6.  s  =  16.1  X  9'=  1304.1  feet  (Aw). 


Page  36. 

Exercise  1.  The  ball  rose  2  seconds  and  fell  2  seconds. 

16.1  X  2s  =  64.4  feet  high  (Ans). 

32.2  X  2  =  64.4  feet  per  second,  velocity  given  (Ans). 

2.  257.6  -=-  32.2  =  8,  the  number  of  seconds  required  for 
gravity  to  stop  it. 

16.1  X  8s  =  1030.4  feet  high  (Aw). 


Page  39. 

Exercise  1.  The  square  of  %  is  %;  therefore  (converse  of 
law  1),  a  pendulum  to  beat  half  seconds  is  one-fourth  of 
39,  or  9|  inches  long.  Similarly,  to  beat  one-third  seconds, 
it  must  be  the  square  of  £,  or  £  of  39,  which  is  4^  inches 
long.  To  vibrate  in  two  seconds  it  is  2*  or  4  times  39 
inches  =  156  inches  long.  To  vibrate  in  one  minute  it  is 
60'  or  3600  times  39  inches  =  140,400  inches  =  11,700  feet 
=  2|f  miles. 

2.  208^  feet  =  2502.4  inches. 

2502.4  H-  39.1  =  64,  which  (law  2)  is  the  square  of  the 
number  of  seconds.  |/64  =  8  seconds  (Ans). 


KEF  TO   NATURAL   PHILOSOPHY. 


Page  41. 

Questions. 

Kind.                                            Fulcrum. 

Balance. 

First. 

Point  of  support. 

Sec-saw. 

First. 

Point  of  support. 

Scissors. 

First. 

Pivot. 

Ladder. 

Third. 

Ground. 

Forearm. 

Third. 

Elbow. 

Tongs. 

Third. 

Joint. 

Pincers. 

First. 

Joint. 

Wheelbarrow. 

Second. 

Axle. 

Shears. 

Third. 

Upper  end. 

Pump-handle. 

First. 

Pivot. 

Claw-hammer. 

First. 

Bearing  of  hammer. 

Eudder. 

First. 

Post. 

Page  43. 

Exercises.  These  may  mostly  be  solved  from  the  simple 
proportion  given  in  Art.  105,  by  transposing  and  substi- 
tuting the  values  given  for  p,  w,  &c. 

1.  2  inches  :  10  inches  : :  20  pounds  :  w,  whence  w  —  100 
pounds  (Ans.~). 

2.  Here  p  -f-  w  :  p  : :  be  (ab  -\-  be)  :  ac,  or  440  :  40  : :  22 
inches  :  ac,  whence  ac  =  2  inches  (Ans.). 

3.  12  inches  :  3  inches  : :  40  pounds  :  10  pounds.     There- 
fore p  =  10  pounds  (J.ns.). 

4.  Leverage  =  240  -h  16  •=  15  (4ns.). 

5.  20  :  4  : :  75  pounds  :  15  pounds  (Ans."). 

6.  300  pounds  :  100  pounds  : :  24  inches  :  8  inches.    There- 
fore ac  =  8  inches,  and  6c  =  24 —  8,  or  16  inches  (Ans.), 

7.  1  inch  :  12  inches  : :  20  pounds  :  240  pounds  (Ans.~). 

8.  2  inches  :  8  inches  : :  3  pounds  :  12  pounds,  p.    By  Art. 
107,  2  :  8  : :  3  feet  :  12  feet  per  second,  the  rate  at  which 
c  moves. 


10  KEF  TO   NATURAL   PHILOSOPHY. 


Pages  46,  47. 

Exercises.     The  general  formula  is  E  X  P=r  X  W. 

1.  20  X  200  =  5  X  W. 

W=  800. 

2.  20X100  =  ;-  X  1000. 

r  =  2. 

3.  20  X  P=i  X  500. 

P=l2l 

4.  R  X  40  =  J  X  1000. 

E  =  121 

5.  Here  1000  X  2  =  moment  of  weight  =  moment  of  the 

2000 

four  men;  hence  —  ^—  =  333.33  =  force  of  the  four  men, 
o 

and  333.33  -=-  4  =  83.33,  force  of  one  man. 

6.  Each  wheel  is  1  foot  in  radius,  and  its  pinion  or  axle 
is  2  inches  in  radius;  therefore  the  power  is  increased  six- 
fold by  each  wheel  and  axle,  or  6  X  6  X  6  =  216  times,  by 
the  train  of  wheels.     216  times  5  pounds  =  1080  pounds 
(Ant). 

7.  10X50  =  10XT7. 
W=  50  kilograms. 

50  kilograms  :  10  kilograms  :  :  20  metres  :  4  metres,  dis- 
tance W  is  hoisted. 


Pages  49,  SO. 

Exercise  1.  There  are  two  movable  pulleys,  each  one 
doubling  the  power.  Hence  20  X  4  —  80  pounds  can  be 
lifted. 

2.  The  weight  will  move  one-fourth  as  far  as  the  power. 

30-4-4  =  7^  feet. 

3.  The  power  required  will  be  one-fourth  of  a  kilogram, 
and  it  will  move  through  four  metres. 

4.  The  power  being  increased  12  times,  and  each  pulley 
increasing  it  two  times,  there  must  be  6  pulleys. 


KEY  TO   NATURAL   PHILOSOPHY.  \\ 

5.  30  pounds  applied  at  b  will  be  equivalent  to  30  X  4  = 
120  at  c.    The  first  pulley  is  fixed,  and  gives  no  new  power; 
the  second  is  movable,  and  doubles  the  power. 

120  X  2  =  240. 

6.  The  wheel  ac  will  have  to  turn  as  many  times  as  to 
cause  a  point  of  its  circumference  to  go  through  6  feet  to 
raise  w  through  3  feet.     Its  radius  is  2  foot.     Hence  its 
circumference  is  3.1416  feet.     Hence  it  has  to  make  nearly 
2  turns. 

7.  At  each  turn  the  larger  sheave  gains  1  link  on  the 
smaller  and  raises  the  movable  pulley  \  link.    By  Art.  107, 
P:  TF::|;31  ::  1  :  62. 

P  =  10  pounds,  therefore  W=  620  pounds  (Ans.). 

8.  P  :  W : :  | :  20  : :  1  :  40. 

W—  3000  pounds,  therefore  P=  75  pounds  (Ans.). 

Page  51. 

Exercise  1.  If  DF  =  20  and  EF  ==  4,  DE  =  1/20'  -f  4'  = 
20.4. 

W:  P::DE  :  EF  : :  20.4  :  4. 

Or  20.4  :  4  : :  500  :  P.     Whence  P=  98+  Ibs.  (Ans.~). 

2.  Here  4  :  20.4  : :  250  :  W.    Whence  W=  1275  Ibs.  (Ans.}. 

3.  If  the  boy  can  push  but  100  pounds  and  must  roll  the 
500-pound  barrel,  the  length  of  the  plane  (skids)  must  be 
five  times  the  height.     5  times  4  feet  =  20  feet  (Ans.). 

4.  As  in  the  last  case,  the  length  of  the  plane  must  be 
990,000  -j-  15,000  ==  66  times  the  height,  or  the  grade  1  in 
66,  which  is  80  feet  per  mile  (Ans.}. 

5.  In  the  lever  CB  the  power  is  increased  five  times,  in  the 
movable  pulley  twice,  and  in  the  inclined  plane  as  20  :  8,  or  ^. 

Hence  a  power  will  lift  a  weight  5  X  2  X  f  =  25  times 
itself.  A  power  of  100  being  employed,  it  will  lift  a  weight 
of  2500. 

6.  To  lift  a  ton,  or  2000  pounds,  will  be  needed  ^  of  2000 
pounds,  or  80  pounds. 

7.  Through  25  feet. 


12  KEY  TO  NATURAL  PHILOSOPHY. 


Pages  55,  56. 

Exercise  1.  Since  the  second-hand  goes  around  60  times 
to  the  minute-hand  once,  if  they  were  the  same  length  the 
end  of  the  second-hand  would  go  sixty  times  as  fast  as  the 
end  of  the  minute-hand.  But,  as  the  hand  is  twice  as  long, 
its  end  has  to  go  twice  as  far.  Hence  its  end  goes  only  % 
of  60,  or  30  times  as  fast. 

2.  The  formula  for  the  space  in  terms  of  the  time  is  s  = 

fcf. 

In  four  seconds  this  would  be 

8  =  J -  x  322  X  16  =  257.6  feet. 
In  five  seconds, 

s=J  X322X  25  =  402.5. 

Hence  in  the  fifth  second  it  would  fall  402.5  —  257.6  — 
144.9  feet. 

3.  The  formula  for  velocity  in  terms  of  the  time  is  v  =  gt. 
At  the  end  of  15  seconds  this  would  be  v  =  3"2.2  X  15  = 
483.    483  feet  per  second  would  be  483  X  3600  =  1,738,800 
feet  per  hour.     1,738,800  -3-  5280  =  329+  miles  per  hour. 

4«  =  i^«.    s  =  iX32x(f>2  =  100. 
Note. — In  some  examples  g  will  be  taken  32,  instead  of 
32.2,  for  convenience. 

5.  The  trains  going  in  opposite  directions,  a  point  in  one 
will  pass  a  point  in  the  other  at  the  rate  of  40  -f  20  =  60 
miles  per  hour,  which  is  one  mile,  or  1760  yards,  per  min- 
ute.    Hence  to  pass  over  one  yard  will  require  Y^SV  °^  a 
minute,  and  110  yards  -jV^  —  iV  minute.     For  the  whole 
express-train  to  pass  would  require  the  front  to  go  the 
length  of  the  train,  66  yards  farther.    Hence  the  whole  time 
required  would  be  -J^  -(-  i  f  |  o  —  iV  m>nute- 

6.  The  steamer  could  go  through  still  water  at  the  rate 
of  6  -|-  4  =  10  miles  per  hour.     Hence  when  it  goes  down 
,lt  will  move  10  +  4=  14  miles  per  hour. 

7.  From  our  formula  for  falling  bodies,  which  can  be 


KEY  TO  NATURAL   PHILOSOPHY.  13 

applied  here  by  putting  </  =  8,  the  acceleration,  we  find 
s  =  %gt2=100. 

8.  A  cylinder  on  its  flat 
end,   if   it    is   upset,    will 
have   to   raise   its    centre 
of  gravity  ;  hence  it  is  in 
stable  equilibrium.     On  its 
curved  surface  the  centre 
of  gravity  will  move  hori- 
zontally, and  have  no  ten- 
dency to  upset  or  replace 
the     cylinder    if    moved. 
Hence    we    have    neutral 
equilibrium. 

9.  The  middle  of  the  opposite  side.     For  the  two  sides 
of  the  board  will  balance  each  other  about  a  line  downward 
from  the  point  of  suspension. 

10.  Since  one  arm  must  be  four  times  as  long  as  the  other, 
one  must  be  one-fifth  and  the  other  four-fifths  the  length 
of  the  lever.     The  fulcrum  will  be  one  foot  from  the  end 
where  the  8  Ibs.  are  placed. 

11.  Let  x  be  the  weight  required  in  the  first  case;  then 

30  X  16  =  15  Xx. 

x  =  32. 
In  the  second, 

x  X  16  =  15  X  30. 
x  =  28\. 

12.  Let  AB  be  the  lever,  C  the  point  where  it  balances, 
and  D  its  middle  point. 


3  A        1       A  2 

Since  it  balances  at  C,  its  centre  of  gravity  must  be  there. 
When  we  place  a  fulcrum  at  D,  the  moment  of  the  lever 
will  be  its  weight  X  CD.  To  balance  this  we  have  a 
weight  with  a  moment  =  3  X  AD  =  9.  Hence  weight  of 
lever  x  1  —  9. 

Weight  =  9. 
2 


14  KEY  TO  NATURAL   PHILOSOPHY. 

13.  Since  the  lever  arms  are  f  and  f  the  length  of  the 
lever,  the  men  will  have  to  bear  f  and  f  of  the  weight 
respectively,  or  120  pounds  (A)  and  80  pounds  (B). 

14.  Since   the   body  appears  to  weigh    1T^  pounds,   it 
must  be  because  the  arm  on  which  it  is  weighed  is  too 

long- 

Let  us  suppose  the  balance  arm  AB  to  be  a  feet  long. 

A  _  (a  —  x)  _  C  _  *  _  B 

A 

Let  BC  =  x  ;  then  AC  =  a  —  x.     Now,  1  pound  at  A  bal- 
ances 1-jig-  pounds  at  B,  or  1  X  (o>  —  x)  = 


x  =  %%a,  instead  of  \a 

as  it  should  be.    Hence  the  other  arm  would  be  ^a,  and 
the  pi-oportion  of  their  lengths  is  16  :  17. 

15.  When  placed  on  the  short  arm,  if  w  represent  the 
weight  of  the  body,  a  the  length  of  the  balance,  and  x  the 
short  arm,  we  have 


A 

4  x  =  w  (a  —  x). 
When  placed  on  the  long  arm, 

9  (a  —  x)  =  wx. 

wa 
From  the  first,  x  =  ^,wm 

Substitute  this  in  the  second, 


or,  36  =  w1. 

w  =  6. 

16.  The  moment  of  each  man's  force  will  be  100  X  5  = 
500,  and  of  the  four  men  2000.     The  lever  arm  of  the 
resistance  is  1  foot  ;  hence  there  is  a  resistance  of  2JtAQ.  = 
2000  pounds. 

17.  In  three  revolutions  the  wheel  will  have  travelled 


KEF  TO  NATURAL   PHILOSOPHY.  15 

through  60  X  3  =  180  inches  =  15  feet.  Since  the  bucket 
rises  one  foot,  the  work  done  on  the  bucket  is  its  weight, 
30  X  1  =  30  foot-pounds.  To  do  this  work  by  a  force  acting 
through  15  feet  will  require  30  -=-  15  =  2  pounds. 

18.  The  length  of  the  plane   will  be  \/  52  -4-  12*  — 13. 
From  proportion  on  page  56,  Philosophy, 

Power  :  65  : :  5  :  13. 
Hence  power  is  25. 

19.  The  circumference  described  by  the  power  will  be 
the  diameter  2  X  3.1416.    Hence,  from  the  law  of  the  screw, 

Weight  :  Power  : :  2  X  3.1416  :  ^  : :  62.832  :  1. 

20.  Given  time  =  ^f  of  a  second.     Required  time  =.  1 
second.     The  converse  of  the  second  law,  page  38,  gives 
the  proportion. 

Given  length  :  Eequired  length  : :  (if)2  :  I1  : :  |f  $  :  1. 
Or  the  pendulum  must  be  lengthened  from  J-|^  to  i-||, 
which  requires  an  addition  of  1^j  of  its  original  length. 

21.  At  4000  miles  above  the  earth's  surface  the  distance 
from  the  centre  would  be  doubled  and  the  intensity  of  grav- 
ity would  be  ^  of  surface  intensity  (Kule,  Art.  41).    Hence 
(third  law,  page  38)  the  time  would  be  2  seconds  (Ans.). 

At  2000  miles  under  ground  the  intensity  of  gravity 
(Art.  41)  is  ^  the  surface  intensity.  The  square  root  of 
£  is  .707  -f- ;  hence  the  time  is  y^-  of  a  second,  or  1.41 
seconds  (Ans.). 

22.  The  time  of  one  vibration  is  ^  of  a  minute  =  1£ 
seconds.     By  law  2,  page  38, — 

I1 :  (1£)2  : :  39.1  inches  :  87.975  inches  (Ans.). 

23.  The  time  of  one  vibration  is  -^  of  a  minute  =  f£ 
seconds :  therefore  (converse  of  law  3,  page  38), — 

(H)2  :  I2  ::  *tt  :  l- 

fit  of  32.2  =  30.1—  (Ans.-). 

Page  67. 

Exercise  1.  22 :  122  : :  5  pounds  :  180  pounds  (Ans.). 
2.  I2 :  121 : :  50  pounds  :  7200  pounds  (Ans.). 


16  KEY  TO  NATURAL   PHILOSOPHY. 

3.  The  conditions  are  the  same  as  in  the  last  example, 
except  that  the  50  pounds  is  applied  to  the  lever  with 
a  leverage  of  6  (ratio  of  3  feet  to  6  inches).     Therefore 
the  weight  lifted  will  be  6  times  7200  pounds,  or  43,200 
pounds  (Ans.). 

4.  One  peck  =  £  of  2150.4,  or  537.6  cubic  inches.     One 
cubic  foot  of  water  weighs  62£  pounds :  therefore  one  cubic 
inch  weighs  17*2g,  and  537.6  cubic  inches  weigh  ^^-|  of 
62£  pounds  =  19£  pounds  (Am.). 

5.  Each  cubic  foot  would  contain  62|-  pounds  of  water. 

6.  The  bottom  supports  the  weight  of  the  water,  and 
each  of  the  four  sides  one-half  the  weight  (Art.  140). 

Pages  78,  79. 

Exercise  1.  900  -=- 1000  =  .900  (Ans). 

2.  The  water  weighs  300  —  200  =  100  grams.    The  acid 
weighs  320  —  200  =  120  grams.      Therefore  the  specific 
gravity  of  the  acid  is  120  -f- 100  =  1.20  (Ans.). 

3.  The  water  buoys  up  the  weight  of  40  —  24  =  16  nails. 
The  lye  buoys  up  the  weight  of  40  — 18  =  22  nails.    There- 
fore the  lye  is  ^f  or  If  times  as  heavy  as  the  water. 

4.  That  the  water  may  exert  its  full  effect  in  supporting 
them. 

5.  It  requires  11£  inches  of  ether  to  weigh  as  much  as 
8  inches  of  water :  therefore  the  specific  gravity  of  ether 
is8-5-ll|  =  .719  (Am.). 

6.  For  the  same  reason  as  above,  the  specific  gravity  of 
the  syrup  is  8  -=-  6.4  =  1.25  (Ans.). 

7.  The  7  ounces  of  iron  displace  one  ounce  of  water: 
therefore  the  specific  gravity  of  the  iron  is  7  -=-  1  =  7 
04ns.). 

8.  The  egg  displaces  260  —  220  =  40  cubic  centimetres 
of  water,  which  weighs  40  grams  (Art.  26).     But  the  egg 
weighs  44  grams.     Hence  its  specific  gravity  is  44  -j-  40  = 
1.1  (4ns.). 


KEY  TO  NATURAL   PHILOSOPHY.  17 

9.  The  dam  is  12  feet  in  perpendicular  height,  and  15 
feet  in  slant  height  :  therefore  its  thickness  at  the  base  must 
be  V  15"  —  12*  =  9  metres.    According  to  Art.  140,  it  there- 
fore supports  half  a  column  of  water  12  by  9  by  1000  metres 
=  54,000  cubic  metres,  and  each  cubic  metre  weighs  1000 
kilograms,  which  gives  54,000,000  kilograms  (/Ins.). 

10.  From  Art.   145  we  learn  that  the  earth  curves  8 
inches,  or  f  of  a  foot,  in  1  mile  and  increases  as  the  square 
of  the  number  of  miles.     Hence  200  feet  divided  by  $  of 
a  foot  gives  the  square  of  the  number  of  miles. 

200  -f-  §  =  300.     1/300  =  17.32  (Ans.). 

Similarly,  from  the  mast  the  water  surface  could  be  seen 
12.24  miles,  for  100  -f-  f  =  150.  1/150  =  12.24,  which 
added  to  17.32  gives  29.56  miles  (Ans.). 

11.  See  Art.  147,  III. 

12.  2^-  —  1  =  1£,  weight  of  the  stone  in  water  compared 

with  an  equal  bulk  of  water.  Therefore  gf  of  the  real  weight 
is  what  the  boy  can  lift.     And  if  120  pounds  =  gf  of  the 

weight  he  can  lift,  that  weight  is  200  pounds  (Ans.}. 

13.  1100  grams  —  975  grams  =  125  grams. 

1100  -=-  125  =  8.8  (Ans.~). 

14.  The  wood  just  buoys  up  the  2  ounces  of  lead.   There- 

fore specific  gravity  =  g--g  =  &  (Ans.~). 


15.  The  keys  displace  6  X  3  X  1  =  4  }  cubic  inches  of 
the  water,  which  is  the  volume  of  the  keys.     The  volume 
of  the  hand  is  6  X  3  X  £  =  9  cubic  inches. 

16.  i|  in  cork  =  ^  in  water. 
^  in  cork  =  yf^  in  water. 
-5^5-  in  cork  =  ^  in  water. 

If  we  divide  the  weight  of  ^  inch  cork  by  the  weight 
of  -^  inch  water,  we  shall  have  the  specific  gravity  of  the 
cork,  or 


2* 


18  KET  TO  NATURAL   PHILOSOPHY. 

17.  As  the  specific  gravity  of  the  body  is  17  times  as 
much  as  the  specific  gravity  of  water,  the  volume  of  a 
certain  quantity  of  it  is  '^  of  the  volume  of  the  same 
quantity  of  water.     The  volume  of  89  ounces  of  water  is 
yflfo.  of  1728  cubic  inches  (Art.  138),  and  J?  of  yflfo  of 
1728  cubic  inches  =  9^1^  cubic  inches  (Ans.~). 

18.  16  ounces  —  6  ounces  =  10  ounces,  weight  of  water. 
14J  ounces  —  6  ounces  =  8  J  ounces,  weight  of  coal  oil. 

And  81  -s-  10  =  .875  (Ans.). 

19.  In  each  case  the  weight  of  the  displaced  liquid  is 
equal  to  the  weight  of  the   hydrometer.      Therefore  we 
have 

11  cubic  inches  oil  =  9  cubic  inches  water. 

1  cubic  inch  oil  =  ^-  cubic  inches  water. 
9  cubic  inches  oil  =  fi  cubic  inches  water. 

And  H  -  9  ==;&=:. Sift- (Ans.> 

20.  As  in  Ex.  19, 

7800  cubic  feet  sea-water  =  8000  cubic  feet  fresh  water. 
1  cubic  foot  sea-water  =  f  £$$  cubic  feet  fresh  water. 
8000  cubic  feet  sea-water  =  94-qfflgM.  CUDie  feet  fresh  water. 
And  M..O|O.<ULO  _*_  gOOO  =  1.026—  (Ans.~). 

As  the  boat  displaces  its  own  weight  of  water,  it  weighs 
62|  pounds  X  8000  =  250  tons  (Ans.). 

21.  Each  cubic  inch  of  cork  buoys  up  itself  and  a  weight 
equal  to  ^j-of  a  cubic  inch  of  water,  or  -ffo  of  i-f|°-  ounces 
—  ]&T  ounces.     And  if  one  cubic  inch  of  cork  buoys  up 
•ffy  ounces,  to  buoy  up  5  ounces  will  take  5  -f-  •£&=  HIT 
cubic  inches  of  cork  (Ans.). 

11^,  cubic  inches  of  cork  weigh  j^|  X  HIT  X -24  = 
lj-£  ounces  (Ans.}. 

22.  2990  grains  — 2960  grains  =  30  grains,  the  weight 
of  the  chalk  in  water.     50  grains  —  30  grains  =  20  grains, 
its  loss  of  weight  in  water.     Therefore  (Art.  152)  50  -h  20 
=  2.5  (vlns.). 


KEY  TO  NATURAL   PHILOSOPHY. 


Pages  89,  9O. 

Exercise  1.  From  Art.  90  v  —  gt,  and  s  —  ^gf. 

By  the  first  formula  t  =  -.     Substituting  this  value  of  t 

in  the  second  formula  we  find  v  =  \/2gs. 
Using  this  formula  we  have  — 

1.  v  =  \/2  x  32  feet  X  1  =  8+  feet  (Ans.}. 

2.  v  =  \/2  X  32.2  feet  X  2  =  11.3+  feet  (4ns.). 

3.  v  =  \/2~xT32.2  feet  X  3  =  13.9+  feet  (Ans.}. 

4.  v  =  V2  x~32^Tfeet  X  4  =  16+  feet  (Ans.). 

5.  v  =  V2X32.2  feet  X  5  =  17.9+  feet  (Ans.}. 

2.  v  =V2~X32T2  feet  X  10  =  25.3+  feet  (Ans.}. 
v  =  V2~X~32.2  feet  X  20  =  35.9+  feet  (Ans.}. 

3.  From  s  =  $gt*,  and  using  answers  to  Ex.  2, 

<=     /*£  =  J2X^  =  1.11  seconds. 
\g         \    32.2 

.  •  .  the  range  =  25.3  feet  X  1.1  1  =  28.083  feet  (Ans.}. 

Again,  t  =  A  /2><t10  =  .78  second. 

^/    32.  2 

.  •  .  the  range  =  35.9  feet  X  -78  =  28.002  feet  (Ans.}. 
These  two  answers  would  agree  exactly  if  the  decimals  had 
been  carried  out  far  enough  all  through  the  solution  (see 
page  81). 

4.  Solved  in  the  book. 

5.  |  of  55f  }  =  34ff  (Ans.}. 

6.  From  the  upper  opening  a  cylindrical  stream  1  inch 
in  diameter  and  60  limes  25.3  feet  will  flow  out.    Reducing 
the  length  to  inches,  and  applying  the  rule  for  finding  the 
volume  of  a  cylinder,  we  have  25.3  X  12  X  60  X  t  X  3.1416 
=  14,306.8+  cubic  inches,  or  61.9  gallons  (Ans.}. 

And  in  the  case  of  the  lower  opening  we  have  35.9  X 
12  X  60  X  4  X  3.1416  =  20,301+  cubic  inches,  or  87.9  gal- 
lons (Ans.}. 

7.  I  of  61.9  gallons  =  38.7  gallons  (Ans.}.  * 
f  of  87.9  gallons  =  54.9  gallons  (Ans.}. 


20  KEY   TO   NATURAL   PHILOSOPHY. 

8.  At  the  sides  of  a  stream  the  water  is  retarded  by  the 
friction  of  the  water  against  the  banks. 

9.  In  a  flood  the  ordinary  body  of  water  in  the  stream 
is  the  bed  of  the  extra  water,  which  rushes  over  it  with 
very  little  retardation  from  friction. 

10.  62£  X  1800  =  112,500  pounds  of  water  per  minute. 
112,500  x  12  =  1,350,000  foot-pounds  per  minute,  the 

potential  energy  of  the  water. 

75  %  of  1,350,000  =  1,012,500  foot-pounds  per  minute 
utilized  by  the  overshot  wheel  (Art.  161). 

1,012,500  -=-  33,000  =  30.7  horse-power  (Ans). 

11.  90  cubic  feet  per  second  =  90  X  60  =  5400  cubic  feet 
per  minute. 

62|  x  5400  =  337,500  pounds  of  water. 
337,500  X  5  =  1,687,500  foot-pounds. 
65  %  of  1,687,500  =  1,096,875  (Art.  162). 
1,096,875  -4-  33,000  ==  33.2  horse-power  (Ans.~). 

12.  8  X  2500  =  20,000  pounds  per  minute. 
20,000  X  18  =  360,000  foot-pounds  per  minute 

80  %  of  360,000  =  288,000  foot-pounds  available  (Art. 
164). 
288,000  -=-  33,000  =  8.7  horse-power  (Ans). 

Page  92. 

No ;  because  the  pressure  of  the  30  inches  of  mercury  in 
the  long  arm  will  raise  the  mercury  above  a  in  the  short 
arm. 

If  the  long  column  is  30  inches  higher  than  the  short  one, 
there  is  a  pressure  of  one  atmosphere  of  mercury  upon  the 
air  in  the  short  tube,  and  the  air  must  be  compressed  to 
half  its  former  bulk.  The  short  column  must  therefore 
have  risen  3  inches,  and,  to  be  30  inches  higher,  the  high 
column  must  be  33  inches  above  c. 

There  are,  then,  3  inches  more  of  mercury  in  the  short 
^arm,  and  30  -{-  3,  or  33,  inches  more  in  the  long  arm. 
Therefore  36  inches  must  be  poured  in. 


KEY  TO  NATURAL  PHILOSOPHY.  21 

When  the  mercury  in  one  tube  is  60  inches  higher  than 
that  in  the  other,  there  is  the  pressure  of  three  atmospheres 
upon  the  air  in  the  short  arm :  therefore  there  are  64  inches 
of  mercury  above  c,  and  64  -(-  4,  or  68,  inches  have  been 
poured  in. 

Pages  114,  115. 

1.  There  is  a  pressure  of  four  atmospheres :  therefore  the 
air  is  compressed  into  the  space  of  i  of  its  original  bulk  of 
6  inches,  or  li  inches  (J.ns.). 

2.  6  —  1J  =  4  J.     And  4£  -f  4£  +  90  =  99  (Ans.). 

3.  Solved  in  the  book. 

4.  Since  the  pressure  is  1£,  or  f,  as  great  as  before,  the 
air  will  occupy  I  as  much  space,  or  4£  inches  (Ans.). 

And  when  one  column  is  45  inches  higher  than  the  other, 
the  pressure  is  2£,  or  -|,  as  great  as  before,  and  the  air  will 
occupy  ^  as  much  space,  or  2^  inches  (,4ns.). 

5.  One  inch   of  mercury  will   support   13.6  -f-  .8,   or   17 
inches  of  alcohol.    And  30  inches  of  mercury,  or  the  weight 
of  the  atmosphere,  will  support  17  inches  X  30  =  42  feet  6 
inches  (Ans.^). 

6.  13.6-=-1.8  =  7f 

And  7|  inches  X  30  =  18  feet  lOf  inches  (Ans.). 

7.  The  pressure  must  be  twice  as  great  as  that  of  the 
atmosphere,  or  there  must  be  one  atmosphere  of  water, — 
f.e.,  34  feet  (J.ns.). 

It  is  the  upper  surface  of  the  water  in  the  tumbler,  or 
the  middle  of  the  tumbler,  that  is  34  feet  deep  (Art.  134). 

8.  There  is  a  pressure  of  three  atmospheres,  one  of  air 
and  two  of  water.     Therefore  we  have  34  feet  X  2  =  68 
feet  (Ans.~). 

And  in  the  second  case,  as  there  are  one  atmosphere  of 
air  and  four  atmospheres  of  water,  the  depth  must  be  34 
feet  X  4  =  136  feet  (Ans.~). 

9.  According  to  Art.  172,  the  mercury  falls  1  inch  for 
every  900  feet  of  ascent.     Therefore  the  barometer  would 


22  KEY  TO  NATURAL  PHILOSOPHY. 

stand  2  inches  lower  than  at  the  level  of  the  sea,  or  usually 
at  about  28  inches. 

10.  As  the  pressure  of  the  atmosphere  is  usually  equal  to 
that  of  30  inches  of  mercury,  when  the  gauge  is  3  inches 
higher  on  one  side  than  on  the  other,  the  pressure,  or  ten- 
sion, of  the  air  in  the  pump  is  only  ^j-,  or  y1^,  of  the  normal 
pressure:  therefore  there  must  be  only  ^  of  the  air  left  in 
the  receiver. 

"When  one  side  is  J  inch  higher  than  the  other,  there 
must  be  -^  =  g1^  of  the  air  left  in  the  receiver. 

11.  If  TsVfr  of  the  air  is  left  in  the  receiver,  it  will  support 
15100  of  30  inches  of  mercury,  or  -fa  inch  (A/is.). 

12.  The  force  to  be  overcome  is  the  pressure  of  the  air 
upon  a  circle  3  inches  in  diameter.     The  area  of  the  circle 
is  (f  )2  X  3.1416  =  7.0686  square  inches.    And  15  pounds  X 
7.0686  =  106+  pounds  (Am.). 

It  may  be  supposed  that  we  ought  to  find  the  pressure 
on  the  surface  of  a  sphere  3  inches  in  diameter.  It  is  true 
that  this  is  the  whole  pressure  on  the  surface  of  the  hemi- 
spheres, but  the  resistance  to  pulling  one  hemisphere  away 
from  the  other  is  only  the  pressure  on  a  section  of  the 
sphere. 

13.  The  area  of  a  circle  2  feet  in  diameter  is  1224X  3.1416 
=  452.39  square  inches.    And  15  pounds  X  452.39  =  6785.85 
pounds. 

14.  32  X  3.1416  —  28.27+  square  inches. 

And  15  pounds  X  28.27  =  424+  pounds  (Am). 

15.  Not  less  than  34  feet  (Art.  192). 

16.  The  atmosphere  at  Denver  would  sustain  a  column 
of  mercury  24.7  inches  high.    Therefore  it  sustains  a  column 
of  water  24.7  inches  X  13.6  =  28  feet  (nearly). 

17.  The  specific  gravity  of  sea-water  is  1.026  (Appendix). 
Therefore 

25  feet--  1.026  =  24.3+  feet  (Am). 


KEY  TO  NATURAL   PHILOSOPHY.  £3 

Page  129. 

At  82°  sound  travels  1090  +  50  =  1140  feet  per  second 
(Art.  205)  :  therefore  we  have 

1140  feet  -r-  120  =  9|  feet  (Ans.). 
And  in  the  second  case  we  have 

1140  feet  -*-  300  =  3*  feet  (Ans.). 

Page  135. 

As  in  the  questions  on  page  129,  we  have  1120  feet  -r-480 
=  2  feet  4  inches.     And  2  feet  4  inches  -=-4  =  7  inches 
(Am.}. 
1120  feet  -f-  280  =  4  feet.     And  4  feet  ~  4  =  1  foot  (Ana.). 

A  column  of  hydrogen  would  have  to  be  about  four  times 
as  long  as  a  column  of  air  to  be  synchronous  with  a  certain, 
tuning-fork. 

Pages  155,  156. 

1.  Touching  the  call-bell  stops  the  vibrations,  and  there- 
fore stops  the  sound. 

2.  No  sound  could  be  heard  unless  the  sounding  body 
was  connected  with  the  ear  by  a  solid  body,  or  by  several 
solid  bodies  touching  one  another.     (Liquids  are  excluded, 
because  it  is  believed  that   there  are  no  liquids  on  the 
moon.) 

3.  ^  of  60  seconds  =  3|  seconds. 

1090  feet  X  3|  ==  40871  feet  (Ana.). 

4.  ._*.  of  60  seconds  =  3  seconds. 

(1090  feet  +  48  feet)  X  3  =  3414  feet  (Ana.). 

5.  They  will  speak  the  words  as  long  after  he  does  as  it 
will  take  sound  to  go  from  him  to  them.   But  it  will  be  just 
as  long  again  before  he  hears  them,  for  their  sound  must 
travel  back  to  him.     The  problem,  therefore,  is  to  find  how 
long  it  takes  sound  to  travel  120  feet.     In  a  temperature 
of  68°  sound  travels  1090  feet  -f  36  feet,  or  1126  feet,  per 
second.     Therefore  we  have  -tffo  or  ^  seconds  (Ana.). 


24  KEY  TO  NATURAL   PHILOSOPHY. 

6.  1150  feet — 1090  feet  —  60  feet.     And 

32°  +  60°  =  92°  (Ans.). 

7.  1142  feet—  1090  feet  =  52  feet. 
And  32°  +  52°  =  84°  (Ans.). 

8.  -3-Q  Of  eo  seconds  =  2J   seconds,  the  time   that   has 
elapsed.     1090  feet  +  28  feet  =  1118  feet.     And  1118  feet 
X2J  =  2515z  feet,  the  distance  the  sound  has  travelled. 
But  one-half  of  this,  or  12571  feet,  is  the  distance  to  the 
barn. 

9.  The  distance  that  the  sound  travels  is  the  bypothenuse 
of  a  right-angled  triangle  whose  perpendicular  is  300  feet 
and  its  base  400  feet.     The  distance  is,  therefore,  500  feet. 
[Draw  the  figure  of  the  right-angled  triangle,  and  work 
this  out.]      The  time,  therefore,  is  -fffg  or  |||  seconds 
(Ans.). 

10.  If  one  string  vibrates  100  times  per  second,  one  twice 
as  long  will  vibrate  50  times,  and  one  of  the  same  length 
and  four  times  as  heavy  as  the  last  one  would  vibrate  25 
times  per  second  (Ans.).     (Arts.  222-3.) 

11.  If  the  string  is  lengthened,  it  vibrates  slower ;  if  short- 
ened, it  vibrates  faster.    When  the  tension  is  made  greater, 
it  vibrates  faster;  when  made  less,  it  vibrates  slower. 

12.  From  Art.  226  we  learn  to  proceed  as  follows :    15 
inches  X  4  =  5  feet,  the  length  of  one  vibration. 

And  1120  feet -i- 5  feet  =  224,  the  number  of  vibrations 
and  the  answer. 

13.  Mi  vibrates  f  of  264,  or  330,  times  per  second  (Ans.). 
(Art.  241.) 

Sol  vibrates  f  of  264,  or  396,  times  per  second  (Ans.) 
Upper  do  vibrates  twice  264,  or  528,  times  per  second  (Ans). 

14.  2  :•§•::  264  :  re's  vibrations.    Therefore  re's  vibrations 
=  148£  (Ans.). 

2  :  J^- : :  264  :  si's  vibrations.  Therefore  si's  vibrations  = 
247|  (Ans.). 

2  :  | : :  264  :/a's  vibrations.  Therefore  fa's  vibrations  = 
176  (Ans.). 


KEY  TO  NATURAL  PHILOSOPHY.  25 

15.  Multiplying  30  inches  by  the  numbers  in  the  series 
given  in  Art.  240,  we  have  the  lengths  of  the  strings  as 
follows : 

do  re  mi  fa  sol  la  si  do 

30    -  26!       24       22|       20.-    18       16       15 

16.  2  :  f  : :  272  :  vibrations  of  1st  A  below.     Therefore 
1st  A  below  vibrates  226§  times. 

2d  A  below  vibrates  226!  -5-2,  or  113J  times. 

3d  A  below  vibrates  113 J  -*-  2,  or  56  j  times. 

4th  A  below  vibrates  56f  -J-  2,  or  28J  times  (Am.*). 

1  :  |  : :  272  :  vibrations  of  1st  A  above.     Therefore 

1st  A  above  vibrates  453J  times. 

2d  A  above  vibrates  453|  X  2,  or  906|  times. 

3d  A  above  vibrates  906|  X  2,  or  1813J  times. 

4th  A  above  vibrates  1813J  X  2,  or  36261  times  (Ans.~). 

17.  1st  C  above  vibrates  272  X  2,  or  544  times. 
2d  C  above  vibrates  544  X  2,  or  1088  times. 

3d  C  above  vibrates  1088  X  2,  or  2176  times. 
4th  C  above  vibrates  2176  X  2,  or  4352  times. 
And  4352  —  36261  =  725  J  (Ans.). 

18.  |  :  1  : :  216  :  do's,  vibrations.     Therefore   do's  vibra- 
tions are  192  (Ans.*). 

216  -j-  2  =  108,  vibrations  of  re  below. 
|  :  -| : :  216  :  la's  vibrations.    Therefore  la's  vibrations  are 
320  (Ans.~). 

19.  A  little  more  than  eleven  octaves.     Multiply  16  by 
2,  then  that  product  by  2,  and  so  on.     Eleven  multiplica- 
tions will  give  us  a  product  a  little  below  38,000. 

20.  Middle  C  vibrates  272  times  per  second.     G  above 
vibrates  408  times. 

Therefore  408  —  272  ==  136,  first  answer  (Art.  246). 
And  510  —  272  =  238,  second  answer. 

21.  At  the  upper,  or  right-hand,  end  of  the  key -board  the 
pitch  is  higher,  the  vibrations  are  more  numerous,  and  the 
differences,  and  therefore  the  number  of  beats,  are  greater. 

In  the  lower  end  of  the  kej--board  the  beats  are  fewer. 


26 


KEY  TO  NATURAL   PHILOSOPHY. 


Page  167. 

1.  The  wall  is  6  feet  from  the  board,  and,  hence,  3  times 
as  far  from  the  light  as  the  board  is.     The  shadow  will  be 
3  feet  square,  or  contain  9  square  feet. 

2.  An  ellipse. 

3.  The  lamp  is  4  times  as  far  from  the  wall  as  the  candle 
is,  and,  since  light  is  proportional  to  the  squares  of  the  dis- 
tances, it  must  be  16  times  as  bright. 

4.  Light  travels  186,380  miles  per  second.     In  3J  years 
there  are  3J  X  365  X  24  X  60  X  60  =  110.366.000  seconds. 

110,366,000  X  186,380  =  20,569,014,080,000  miles. 

5.  No;    because   they  may 
have  gone  out  of  existence, 
and  the  light  be  still  coming 
to  us. 

6.  About  1^  seconds.     Less 
than  \  second. 

Pages  173,  174. 

1.  When    we    look    into   a 
mirror    perpendicularly,    the 
reflections  from  the  two  faces 
will  coincide  as  AB,  and  we 
will    not    see    double    reflec- 
tion.    If  the  ray  strikes  the 
mirror  a  little   obliquely,  as 
CD,  the   reflected  light 
from   the   two  surfaces 
will    come    out     much 
closer  together  than  if 
it  strikes  very  obliquely, 
as  EF. 

2.  Images  will  come  into  the  eye  from  various  reflec- 
tions from  the  mirrors. 


KEY  TO  NATURAL  PHILOSOPHY. 


27 


3.  The  man  CD  can  see  his  head  reflected  perpendicu- 
larly from  A,  and  his  feet  from  B. 

4.  The  tremors  of  the  atmosphere  A 
make  the   light  from  the   stars  to 
quiver,  and  so  give  them  apparent 
size,  and  the  reflections  in  the  air 
make  the  whole  sky  somewhat  light. 
For  the  same  reason  light  is  reflected 
into   shadows.      We  see   only   rays 
which  enter  the  eye,  but  a  beam  of 
light  going   through   a  dark  room 
lights  up  particles  of  dust,  and  these 
reflect  the  light  to  the  eye. 


Page  217. 

2.  122°  F.  is  90°  above  freezing, 
and  90°  F.  =  50°  C. 

3.  10°  C.  =  18°  F.,  and  18°  above  freezing  is  50°  above 
zero. 

4.  _4Q°  F.  is  72°  below  freezing,  and  72°  F.  =  40°  C. 

5.  Since  772  units  are  required  to  raise  one  pound  through 
1°  F.,  and  1°  F.  =  f°  C.,  to  raise  one  pound  through  1°  C. 
would  require  f  of  772  =  1389.6. 


Page  220. 

Question.  273°  C.  —  491.4°  F.  below  the  freezing-point, 
or  491.4  —  32  =  459.4°  below  zero. 


Page  224. 

1.  Because  the  water  is  dried  out.    This  more  than  makes 
up  for  the  expansion  due  to  heat. 


28  KEY    TO   NATURAL   PHILOSOPHY. 

2.  Because  the  wires  are  expanded  by  heat  and  rendered 
longer. 

3.  Because  the  tire  is  larger  when  hot.     When  it  cools  it 
contracts,  and  binds  the  wheel  together. 

4.  It  will.     Because  some  heat  is  taken  from  the  coffee  to 
dissolve  the  sugar. 

Page  227. 

3.  Since  the  volume  of  a  gas  increases  ^  its  volume  at 
zero  for  each  degree,  for  546°  it  will  expand  fff  its  volume 
at  zero,  or  twice.  Add  this  to  the  original  volume,  and  we 
have  the  new  volume, — three  times  the  old. 

Page  231. 

1.  Stoves  should   not  be   bright,   for  that  makes  them 
poorer  radiators  of  heat.     Teapots  should  be,  for  we  do  not 
wish  them  to  radiate.     The  same  is  true  of  the  cylinders 
of  steam-engi ncs. 

2.  Dark  colors  will  absorb  more  luminous  heat  than  light 
colors  will.    Hence  light  colors  are  cooler  in  sunlight.    But 
by  a  dark  stove  there  is  no  difference. 

3.  Very  little  heat  will  go  through  the  alum,  and  nearly 
all  through  the  rock-salt.     Hence  the  focus  of  the  latter 
will  be  the  hotter. 

4.  Four  times.     There  is  the  same  law  as  with  light  and 
gravitation  and  all  such  forces. 

5.  The  luminous  waves;  for,  being  nearer  the  violet  end, 
it  is  shown  that  thej^  are  refracted  most,  and  hence  must 
bo  more  rapid. 

6.  The  heat  from  an  open  fire  will  penetrate  a  glass  screen 
much  more  readily  than  dark  heat  will :  hence  it  is  less 
effective. 

Page  234. 

1 1.  Because  the  vessel  of  better  conductive  power  more 
quickly  carries  the  heat  through  its  mass,  and  thus  there  is 
not  unequal  expansion  and  contraction  in  its  different  parts. 
2.  Because  these  have  poor  conductive  power. 


KEY  TO   NATURAL   PHILOSOPHY.  29 

3.  Because  it  will  not  so  readily  carry  away  the  heat  of 
the  body. 

4.  Because  there  is  around  his  hand  a  layer  of  vapor  from 
the  perspiration  of  the  skin,  which,  being  a  poor  conductor, 
prevents  for  a  little  time  the  heat  of  the  iron  from  being 
conveyed  to  the  nerves. 

5.  The  brass  rapidly  conducts  the  heat  away,  and  the 
wood  does  not.     Hence  the  heat  accumulates  in  the  latter 
case  and  scorches  the  paper. 

6.  They  surround  the  house  with  a  partition  of  air,  which 
is  not  a  good  conductor  of  heat,  and  so  retain  it  in  the 
house. 

7.  Not  so  effective.   Because  dark  heat  will  not  penetrate 
glass. 

8.  They  reflect  the  heat  back  to  the  earth  and  retain  it 
there. 

9.  They  are  not  always.     If  our  hand  is  cooled  for  any 
reason,  any  object  warmer  than  itself  will  feel  warm  to  it. 
In  the  case  mentioned,  the  tepid  water  will  feel  warm  to 
the  hand  previously  in  the  cold  water,  and  cold  to  the 
other. 

Pages  241,  242. 

1.  113°  F.  is  81°  above  freezing. 

81°  F.  =-f  of  81°  C.  =  45°  C.,  or  45°  above  0°. 
140°  F.  is  108°  above  freezing. 
108°  F.  =f  of  108°  C.  =  60°  C. 

2.  15°  C.=|  of  15°  F.  =  27°  F.  above  freezing,  or  32° 
-f  27°  =  59°  F. 

35°  C.  =f  of  35°  F.  =  63°  F.  above  freezing,  or  32°  -f 
63°  =  95°  F. 

3.  It  would  expand  tfft  of  its  volume  at  0°.    |ff  of  30 
=  11.     30  +  11  =  41  (Art.  366). 

4.  185°  F.  =  153°  above  0°  =  85°  C.     Hence  98  cubic 
inches  =  f^f  +  -^fa  times  the  volume  at  0°.     If  98  is  f  ff , 
the  volume  at  0°  will  be  f|f  of  98,  and  the  volume  at  10° 


30  KEY  TO  NATURAL   PHILOSOPHY. 


will  be  |^|  -j-  JT%  =  |ff  times  the  volume  at  0°.  Therefore 
the  volume  at  10°  =  fff  of  f£f  of  98  =  77+  cubic  inches 
(Am). 

5.  If  50  is  fff  the  volume  at  0°,  then  the  volume  at  0° 
is  !£|  x  50,  and  at  15  the  volume  would  be  f|f  X  50  X  fff 
=  53.8  nearly. 

6.  Since  it  expands  ^rs  f°r  eacn  degree,  the  expansion 
of  20  cubic  inches,  or  £  of  its  volume,  at  0°  must  represent 
i.-j-^i^  or  -2-p  —  546°  of  temperature. 

7.  The  pound  of  water  if  lowered  to  0°  C.  will  lose  50 
units  of  heat,  and  this  is  given  to  the  ice.    Now,  it  requires 
80  units  of  heat  to  melt  1  pound  of  ice  (Par.  368).     The 
50  units  will  melt  f  £  of  a  pound,  or  10  ounces. 

8.  The  3  pounds  at  12°  will  have  36  units  of  heat,  and 
the  3  pounds  at  16°  will  have  48  units;  in  all,  84  units. 
If  6  pounds  have  84  units,  each  pound  must  be  of  the  tem- 
perature of  14°. 

9.  4  X  7  =  28,  and  6  X  12  =  72. 

72  +  28=  100  units  of  heat  in  the  mixture  of  10  pounds. 
100  -r-  10  =  10,  the  temperature. 

10.  The   water   evaporates   rapidly   through   its  pores. 
This  evaporation  takes  heat  from  the  water,  and  so  keeps 
it  cool. 

11.  The  evaporation,  or  conversion  into  a  gas,  is  so  rapid 
that  heat  is  taken  from  the  adjacent  particles,  and  they 
freeze. 

12.  If  the  specific  heat  of  iron  is  .1,  it  takes  -^  unit  to 
raise  1  pound  through  1°.     The  iron  therefore  loses  10°  to 
give  the  water  1°.     They  thus  become  11°  nearer  the  same 
temperature  each  time  the  water  is  raised  1°.     To  agree, 
the  water  will  have  to  be  raised  through  99  -f-  11  =  9 
degrees. 

13.  The  air  on  a  high  mountain  is  rare  and  dry.     The 
sun-rays  pass  through  it  without  heating,  and  the  radiation 
from  the  earth  is  not  retained  by  the  cloak  of  moisture 
which  exists  in  the  valleys. 


KEY  TO   NATURAL  PHILOSOPHY.  31 

14.  Silver-foil  being  a  poor  absorber,  the  mercury  will 
not  rise  so  fast  (Art.  389). 

15.  One  way  would  be  to  coat  them  all  with  wax  and 
place  one  end  in  a  vessel  of  hot  water.     The  distance  to 
which  the  wax  would  melt  would  determine  the  relative 
conductive  power. 

16.  Platinum  is  a  poor  conductor  of  heat,  and  copper  is 
a  good  one. 

17.  386  X  2000  =  772,000  foot-pounds  of  work  done.    A 
unit  of  heat  is  772  foot-pounds :  hence  1000  units  are  re- 
quired. 

18.  1544  X  68  =  104,992    foot-pounds^of   work.     This 
would  raise  1  pound  of  water  through 

104,992  -=-  772  =  136°,  or  68  pounds  through  2°. 

19.  It  requires  ^  as  much  heat  to  raise  a  pound  of  iron 
as  a  pound  of  water.     To  raise  1  pound  of  water  1°  re- 
quires 772  foot-pounds.     To  raise  it  100°  requires  77,200 
foot-pounds.    To  raise  1  pound  of  iron  through  100°  requires 
7720  foot-pounds.     This  would  lift  7720  pounds,  or  about 
3  z  tons,  through  one  foot. 

Page  279. 

1.  The  fur  will  become  positively  electrified  by  friction 
against  the  boy's  clothing,  and  the  positive  electricity  will 
be  conducted  through  the  hand  to  the  boy  who  holds  it, 
thus  giving  him  a  positive  charge.     The  other  boy  will  at 
the  same  time  be  negatively  electrified  through  his  clothing. 
If  the  two  boys  bring  their  knuckles  near  to  each  other,  a 
spark  will  pass. 

2.  She  will  be  positively  electrified,  just  as  the  prime  con- 
ductor is  positively  electrified  by  presenting  a  row  of  points 
to  the  plate  of  the  machine. 

3.  It  will  be  positively  charged. 

4.  The  end  nearest  the  prime  conductor  will  be  — ,  that 
nearest  the  negative  conductor  will  be  -f-. 

5.  In  the  first  case,  they  should  dance  vigorously  be- 


32  KEY   TO   NATURAL   PHILOSOPHY. 

tween  the  rod  and  the  table.  In  the  second  case,  they 
should  be  attracted  to  the  rod,  then  repelled  to  the  pane 
of  glass,  and  there  remain. 

6.  By  Art.  450,  1  X  -1  X  22  =  4  dynes  (Ans.). 

7.  As  above,  4  X  2  X  I2  =  8  dynes,  attraction  (Am.). 
When  they  touch,  the  two  — units  neutralize  two  of  the 

-f-  units  and  there  are  left  two  -(-units,  which  distribute 
themselves  one  on  each  ball. 

Then  1  X  1  X  I2  =  1  dyne,  repulsion  (Ans.). 

8.  16  (dynes)  =  32(+unit8)4r(~U'"'"). 
Clearing  of  fractions  and  transposing, 

*  32  a;  =  256. 

x  =  8  — units  (Ans). 

9.  By  Art.  61,  1  gram  =  980  dynes.      980  X  5  =4900 
dynes,  repulsion.    490  (-f  units)  X  x  (+  units)  =  4900. 

490  x  =  4900. 

x  =  10  -{-  units  (Ans.~). 

Page  334. 

1.  12  X  24  =  288  ohms,  resistance  of  line  wire. 

1.8  X  12  =  21.6  volts,  E.  M.  F.  of  12  cells. 
By  Ohm's  law  the  current  is 
21.6  -4-  288  =  .075  amperes,  or  75  milliamperes  (Ans.). 

2.  Let  x  =  the  number  of  cells  required. 

288  ohms,  as  above,  =  resistance  of  line  wire. 
104.25  X  4  =  417  ohms,  resistance  of  relays. 
.5  x  =  resistance  of  battery. 
288  -f  417  -\-  .5  x  =  total  resistance  of  the  line. 
1.8  x  =  E.  M.  F.  of  whole  battery. 

Ohm's  law,  cleared  of  fractions  and  transposed,  becomes 
E  =  CE.    Or, 

1.8  x  =  .075  (288  +  417  +  .5  *). 

1. 8  x  =  52.875-1-.  0375  x. 

1.7625  a;  =  52.875. 

x  =  30  ( Ans). 


KEY  TO  NATURAL   PHILOSOPHY.  33 

3.  By  Art.  501,  a  copper  wire  ^  of  an  inch  in  diameter 
has  a  resistance  of  1  ohm  in  962  feet. 

The  half-inch  wire  has  -fa  the  resistance  per  foot  of  the 
one-tenth-inch  wire  (Art.  496). 

Therefore  962X25  =  24,050  feet  of  half  inch  wire  for 
each  ohm  of  resistance. 

5280  X  400  =  2,112,000  feet  in  400  miles. 

2,112,000  -4-  24,050  =  87.8  ohms  (Ans.~). 

3000  -f-  87.8  =  34.17  amperes  (Ans.). 

(By  Art.  495)  34.17  X  3000  =  102,510  watts  =  102.51 
kilowatts  (Ans.). 

4.  Current  in  the  shunt  is  always  f$  =  5  amperes  (.4ns.). 
One  lamp  takes  ^^  =  ^  ampere  (Ans.). 

200  lamps  take  %  X  200  =  100  amperes  (Ans.). 

5.  I  X  600  =  300  amperes  for  lamps. 

300  -f  12.05  +  5  (for  field  magnet)  =  317.05  amperes, 
whole  current  required. 
80  X  317.05  =  25,364  watts. 
(Art.  495)  25,364  -=-  746  =  34  horse-power  (Ans). 


amflORMALSCHOOL, 


319* 


UC  SOUTHERN  REGIONAL  LIBRARY  F 


000  947  352     1 


UNIVERSITY  of  CALIFORNIA 

AT 

L06  ANGELES 
LIBRARY 


